Why is attempt 3 a dead end for black? [#2169]
188.8.131.52: Why is attempt 3 a dead end for black?
(2010-01-27 08:38) [#7131]
I hace been struggling to correctly determine when something can be declares dead for some time. On attemp 3, I see that if black tries to kill white, it loses. But on the other hand, if white tries to kill black, it seems to fail as well, enabling black to get two eyes. So if white can't kill black, why should the black stones be dead?
Suppose black asks white to probe he's dead, white has two choices. One is to play 1-1, then black kills the white group in the only position left. White may try to play 4 again, but then black will reply with a 2-1 (not sure what axis goes first) move. After this, white can't attempt anything else.
A similar result comes from white initially playing 1-3.
Is there a rule that forbids black's claim his group ain't dead as white can't naturally kill it? What am I overlooking?
184.108.40.206: Re: Why is attempt 3 a dead end for black?
(2010-01-27 08:58) [#7132]
You're failing to "eat" because you are trying from the inside.
If white had to probe he can eat he will play on the outside, and there will be only one liberty remaining on the inside (1-3). And if black answers by playing inside he wont do anything else that eating a 3stone-shape which is not enough to live.
220.127.116.11: Re: Why is attempt 3 a dead end for black?
(2010-01-27 09:01) [#7133]
After , the black group is in atari. White will capture at a, but connecting there doesn't help, as white will then capture all the stones with a move at the 1-1 point.
: Re: Why is attempt 3 a dead end for black?
(2010-01-28 00:39) [#7134]
There are cases where neither group can capture the other: it's called seki, and that page has lots of examples, so it may help your understanding. This position isn't a seki, although it kindof looks like it.